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Q.

A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A small capacitor C is then charged from C0, discharged and charged again; the process being repeated n times. Due to this, the potential of the larger capacitor is decreased to V. The value of C is

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a

C0V0V1/n

b

C0V0V1/n−1

c

C0VV0−1n

d

C0VV0n+1

answer is B.

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Detailed Solution

Potential of larger capacitor after the first charging isV1=C0V0C+C0After second charging, potential isV2=C0V1C+C0=C0C+C02V0After nth charging, potential isVn=C0C+C0nV0But  Vn=VSo C=C0V0V1/n−1
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